How to Verify Rolle`s Theorem

Practice Questions With Step-By-Step Solutions

Question 1

Let [f(x)=x^{3}-3x^{2}-x+3] on [0,3]. Show Rolle’s Theorem applies and find [c].

Step-by-step Solution

[f(x)] is a polynomial ⇒ continuous on [(0,3)] and differentiable on [(0,3)].
Compute end values: [f(0)=0^{3}-0-0+3=3], [f(3)=27-27-3+3=0]. So [f(0)=3], [f(3)=0] → not equal.
Since [f(0)\ne f(3)], Rolle’s Theorem does not apply.

Conclusion: Rolle’s Theorem not applicable (end values differ).

Question 2

Let [f(x)=x^{2}-4x+4] on [(1,3)]. Determine whether Rolle’s Theorem applies; if yes find [c].

Step-by-step Solution

Polynomial ⇒ continuous & differentiable on interval.
[f(1)=1-4+4=1], [f(3)=9-12+4=1] ⇒ [f(1)=f(3)]. Condition holds.
Compute derivative: [\dfrac{d}{dx}(f(x))=2x-4]. Set to zero: [2x-4=0 ⇒ x=2].
Check [2\in(1,3)] ⇒ yes.

Conclusion: Rolle’s Theorem applies; one [c=2].

Question 3

Let [f(x)=\sin x] on [(\pi/2,3\pi/2)]. Verify Rolle’s Theorem and find [c].

Step-by-step Solution

[\sin x] is continuous & differentiable everywhere.
[f(\pi/2)=1], [f(3\pi/2)=-1] ⇒ not equal.
Rolle’s Theorem does not apply.

Conclusion: Not applicable because [f(a)\ne f(b)].

Question 4

Let [f(x)=\cos x] on [(0,2\pi)]. Does Rolle’s Theorem apply? Find all [c] in [(0,2\pi)] with derivative zero.

Step-by-step Solution

[\cos x] continuous & differentiable everywhere.
[f(0)=1], [f(2\pi)=1] ⇒ equal.
Derivative: [\dfrac{d}{dx}(\cos x)=-\sin x]. Solve [ -\sin x=0 ⇒ \sin x=0 ].
Solutions in [(0,2\pi)]: [x=\pi]. (Also endpoints [0] and [2π] excluded.)

Conclusion: Rolle’s Theorem applies; at least one [c=\pi]. (Only [\pi] lies in [(0,2\pi)].

Question 5

Let [f(x) = \begin{cases} x+2 & x\le 0\ x^{2}+2 & x>0 \end{cases}] on [(-1,1)]. Determine if Rolle’s Theorem applies.

Step-by-step Solution

Check continuity at 0: left-limit [f(0^{-})=0+2=2], right-limit [f(0^{+})=0^{2}+2=2] ⇒ continuous at 0. Elsewhere piecewise polynomials are continuous. So continuous on [-1,1].
Check differentiability on [(-1,1)]: For [x<0] derivative =1. For [x>0] derivative =2x. At [x=0], left derivative [=1], right derivative =0 ⇒ derivatives not equal ⇒ not differentiable at [0].
Compute endpoints: [f(-1)=-1+2=1], [f(1)=1^{2}+2=3] ⇒ not equal.

Conclusion: Rolle’s Theorem does not apply (fails differentiability at [0] and [f(-1)≠f(1))].

Question 6

Let [f(x)=x^{3}-3x] on [(-\sqrt{3},\sqrt{3})]. Show Rolle’s Theorem applies and find all [c].

Step-by-step Solution

Polynomial ⇒ continuous & differentiable.
Evaluate: [f(-\sqrt{3}) ][= -3\sqrt{3} + 3\sqrt{3} ][= 0], [f(\sqrt{3}) ][= 3\sqrt{3} – 3\sqrt{3} ][= 0] ⇒ equal.
Derivative: [\dfrac{d}{dx}(f) ][= 3x^{2}-3 = 3(x^{2}-1)]. Set zero: [3(x^{2}-1)=0 ⇒ x^{2}=1 ⇒ x=±1].
Check that [±1] lie in [(-√3,√3)], [(√3≈1.732)] ⇒ yes both inside.

Conclusion: Rolle’s Theorem applies; two points satisfy derivative zero: [c=-1] and [c=1].

Question 7

Let [f(x)=\sin x + \cos x] on [(0,\pi)]. Verify conditions; find [c].

Step-by-step Solution

[\sin x + \cos x] continuous & differentiable everywhere.
[f(0)=0+1=1], [f(\pi)=0-1=-1] ⇒ not equal ⇒ Rolle’s Theorem fails.

Conclusion: Not applicable (end values differ).

Question 8

Let [f(x)=x^{4}-2x^{2}] on [(-1,1)]. Determine whether Rolle’s Theorem applies and find c.

Step-by-step Solution

Polynomial ⇒ continuous & differentiable.
[f(-1)=1-2=-1], [f(1)=1-2=-1] ⇒ equal.
Derivative: [\dfrac{d}{dx}f][ = 4x^{3}-4x][ = 4x(x^{2}-1)]. Set zero: [4x(x^{2}-1)=0 ⇒ x=0, x=±1].
Values in open interval [(-1,1)]: only [x=0]. ([x=±1] are endpoints excluded.)

Conclusion: Rolle’s Theorem applies; at least one [c=0].

Question 9

Let $ \displaystyle f(x)=\left\{ \begin{array}{l}\sin x,\text{ }0\le x\le \pi /2\\1,\text{ }\pi /2<x\le \pi \text{ }\end{array} \right.$ on [(0,\pi)]. Does Rolle’s Theorem apply?

Step-by-step Solution

Check continuity at [x=\pi/2]: left-limit [\sin(\pi/2)=1], right-value [1] ⇒ continuous at [π/2]. So continuous on [(0,π)].
Differentiability: on [(0,π/2)] derivative = cos x; on [(π/2,π)] derivative =0. At [x=π/2], left derivative [= cos(π/2)=0], right derivative =0 ⇒ derivatives match ⇒ differentiable at [π/2]. Actually both sides give 0 — so differentiable everywhere on [(0,π)].
Evaluate endpoints: [f(0)=0], [f(π)=1] ⇒ not equal.

Conclusion: Rolle’s Theorem does not apply (end values differ), though function happens to be continuous & differentiable.

Question 10

Let [f(x)=x^{2}\sin(\pi x)] on [(0,2)]. Verify Rolle’s Theorem and find at least one [c].

Step-by-step Solution

Product of continuous differentiable functions ⇒ continuous & differentiable on [(0,2)].
Evaluate endpoints: [f(0)=0^{2}\sin 0 = 0], [f(2)=4\sin(2\pi)=4\cdot 0 =0] ⇒ equal.
Compute derivative using product rule:
[\dfrac{d}{dx}f ][= 2x\sin(\pi x) + x^{2}\cdot \pi\cos(\pi x)].
Set to zero: [2x\sin(\pi x) + \pi x^{2}\cos(\pi x)=0]. Factor [x]: [x(2\sin(\pi x) + \pi x\cos(\pi x))=0].
Solutions: [x=0] (endpoint, excluded) or solve [2\sin(\pi x) + \pi x\cos(\pi x)=0]. This transcendental equation may have solutions in [(0,2)]. Observe that x=1 is a root? Test [x=1]: [2\sin(\pi) + \pi\cdot1\cdot\cos(\pi) = 0 + \pi(-1) = -\pi ≠0]. Try x=0.5: [2\sin(\pi/2)=2\cdot1=2], [\pi(0.5)\cos(\pi/2)=\tfrac{\pi}{2}\cdot 0=0], sum 2 ≠0. Try x=1.0 we saw negative. By continuity the expression changes sign between 0.5 and 1, so there exists at least one root in [(0.5,1)] by Intermediate Value Theorem. So at least one [c\in(0,2)] solves derivative=0. (Exact closed form not elementary.)

Conclusion: Rolle’s Theorem applies; at least one interior point [c] (numerically between 0.5 and 1) satisfies [\dfrac{d}{dx}f(c)=0].